1) AYDZ is cyclic, and easily we find YZ perpendicular to A-symmedian. 2) Lemma - Carnot: the perpendiculars from the vertices of a triangle onto the sides of another one are concurrent iff the perpendiculars from the vertices of the latter onto the sides of the first one are concurrent. The proof is easy, and I do not think we need to prove it here.
Now, to the problem: Since the Perpendiculars from P to BC, from Z to AB and from Y to AC concur at D, we get that the perpendiculars from A to YZ, from B to ZP and from C to PY concur as well, hence Q belongs to A-symmedian.
1) AYDZ is cyclic, and easily we find YZ perpendicular to A-symmedian.
ResponderEliminar2) Lemma - Carnot: the perpendiculars from the vertices of a triangle onto the sides of another one are concurrent iff the perpendiculars from the vertices of the latter onto the sides of the first one are concurrent.
The proof is easy, and I do not think we need to prove it here.
Now, to the problem: Since the Perpendiculars from P to BC, from Z to AB and from Y to AC concur at D, we get that the perpendiculars from A to YZ, from B to ZP and from C to PY concur as well, hence Q belongs to A-symmedian.
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Stan Fulger